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Question: A man wants to distribute 101 coins a rupee each, among his 3 sons with the condition that no one re...

A man wants to distribute 101 coins a rupee each, among his 3 sons with the condition that no one receives more money than the combined total of other two. The number of ways of doing this is :-
A. 100C2−2×52C2{}^{100}{C_2} - 2 \times {}^{52}{C_2}
B. 100C23\dfrac{{{}^{100}{C_2}}}{3}
C. 1275
D. 103C26\dfrac{{{}^{103}{C_2}}}{6}

Explanation

Solution

The sum of the amount with three sons will be equal to 101. We will further apply the condition given that no one receives more money than the combined total of the other two.

Complete step-by-step answer:
Given, the amount is Rs.101Rs.101
Now, let the amount of three sons be xx, yy and zz respectively.
Given condition is x+y+z=101x + y + z = 101 and x⩽y+zx \leqslant y + z
Since,
x+y+z=101 x=101−(y+z)  x + y + z = 101 \\\ x = 101 - (y + z) \\\
Now, x⩽y+zx \leqslant y + z
⇒x⩽101−x ⇒2x⩽101 ⇒x⩽50  \Rightarrow x \leqslant 101 - x \\\ \Rightarrow 2x \leqslant 101 \\\ \Rightarrow x \leqslant 50 \\\
Similarly, y⩽50y \leqslant 50 and z⩽50z \leqslant 50
And, x+y+z=101x + y + z = 101
We will apply the concept of multinomial to find out the number of ways of the above-mentioned condition.
The corresponding Multinomial is (1+x+x2+...x50)3{(1 + x + {x^2} + ...{x^{50}})^3}
Hence, total number of distribution is equivalent to coefficient of x101{x^{101}} in the expansion of (1+x+x2+...x50)3{(1 + x + {x^2} + ...{x^{50}})^3}
Coefficient of x101{x^{101}} in the expansion (1+x+x2+...x50)3{(1 + x + {x^2} + ...{x^{50}})^3}
The expansion (1+x+x2+...x50)3{(1 + x + {x^2} + ...{x^{50}})^3} can be simplified with the help of concept of sum of Geometric progression i.e., Sum =a(1−rn)1−r = \dfrac{{a(1 - {r^n})}}{{1 - r}} where, a is the first term and r is the common ratio.
Applying,
(1+x+x2+...x50)3=(x51−1x−1)3{(1 + x + {x^2} + ...{x^{50}})^3} = {(\dfrac{{{x^{51}} - 1}}{{x - 1}})^3}
The coefficient of x101{x^{101}} in the expansion of (x51−1x−1)3{(\dfrac{{{x^{51}} - 1}}{{x - 1}})^3}
The coefficient of x101{x^{101}} in the expansion of (x51−1)3(1−x)−3{({x^{51}} - 1)^3}{(1 - x)^{ - 3}}
The coefficient of x101{x^{101}} in the expansion of(x153−1−3x102+3x51)(1+3x+4C2x2+5C3x3+...+...)({x^{153}} - 1 - 3{x^{102}} + 3{x^{51}})(1 + 3x + {}^4{C_2}{x^2} + {}^5{C_3}{x^3} + ... + ...)
The coefficient of x101{x^{101}} is 103C101−352C50{}^{103}{C_{101}} - 3{}^{52}{C_{50}}
Hence, the coefficient of x101{x^{101}} is 1275.

So, the correct answer is “Option C”.

Note: In the multinomial theorem, the sum is n1,n2,...,nk{n_{1,}}{n_2},...,{n_k} taken over such that n1+n2+n3+...+nk=nn{}_1 + {n_2} + {n_3} + ... + {n_k} = n. The multinomial theorem gives us a sum of multinomial coefficients multiplied by variables. In other words, it represents an expanded series where each term in it has its own associated multinomial coefficient.